The Four Clues
I am a 4-digit number. My first digit is twice my last. My second digit is my first and last added together. All my digits add up to 20, and I am odd. What number am I?
Write every digit in terms of the last one.
6923
Call the last digit x: the digits are 2x, 3x, 20 - 6x and x. x can only be 1, 2 or 3; 1 breaks the third digit and 2 makes it even, so x = 3: 6923.
How to solve it
Write every digit in terms of the last digit, x.
The first digit is twice the last, so it is 2x. The second is the first plus the last, so 2x + x = 3x. The four digits add to 20, so the third digit is 20 - 2x - 3x - x = 20 - 6x.
Now test values. x cannot be 0, because the first digit would be 0. x cannot be 4 or more, because 3x would be 12 or more, which is not a single digit. That leaves 1, 2 or 3.
If x = 1, the third digit is 20 - 6 = 14, not a digit. If x = 2, the number is 4682, but it ends in 2, so it is even. If x = 3, the digits are 6, 9, 20 - 18 = 2 and 3: the number 6923.
Check: 6 is twice 3, 9 = 6 + 3, 6 + 9 + 2 + 3 = 20, and it ends in 3, so it is odd. The usual slip is stopping at 4682, which meets every clue except being odd.
